A193594 Number of attractors under iteration of sum of cubes of digits in base b.

1, 6, 9, 6, 9, 34, 11, 28, 15, 46, 22, 50, 49, 60, 86, 86, 60, 128, 22, 58, 118, 93, 64, 185, 5, 109, 102, 100, 122, 184, 51, 94, 205, 131, 173, 275, 67, 216, 131, 127, 34, 360, 114, 78, 215, 213, 393, 479, 76, 254, 634, 197, 214, 496, 348, 170, 437, 349, 290

Offset: 2

Author

Martin Renner, Jul 31 2011

Keywords

• nonn
• base

Comments

If b>=2 and a >= 2*b^3, then S(a,3,b)

Examples

			In the decimal system all integers go to (1); (153); (370); (371); (407) or (55, 250,133); (136, 244); (160, 217, 352); (919, 1459) under the iteration of sum of cubes of digits, hence there are five fixed points, two 2-cycles and two 3-cycles. Therefore a(10) = 5 + 2*2 + 2*3 = 15.
		

Links

• H. G. Grundman, E. A. Teeple, Generalized Happy Numbers, Fibonacci Quarterly 39 (2001), nr. 5, p. 462-466.

Crossrefs

Cf. A193586.

Programs

Maple

S:=proc(n, p, b) local Q, k, N, z; Q:=[convert(n, base, b)]; for k from 1 do N:=Q[k]; z:=convert(sum(N['i']^p, 'i'=1..nops(N)), base, b); if not member(z, Q) then Q:=[op(Q), z]; else Q:=[op(Q), z]; break; fi; od; return Q; end:
NumberOfAttractors:=proc(b) local A,i,Q; A:=[]: for i from 1 to 2*b^3 do Q:=S(i,3,b); A:=[op(A),Q[nops(Q)]]; od: return(nops({op(A)})); end:
seq(NumberOfAttractors(b),b=2..20);

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement.